(16-2x)(10-2x)=1

Solution for (16-2x)(10-2x)=1 equation:

(16-2x)(10-2x)=1

We move all terms to the left:

(16-2x)(10-2x)-(1)=0

We add all the numbers together, and all the variables

(-2x+16)(-2x+10)-1=0

We multiply parentheses ..

(+4x^2-20x-32x+160)-1=0

We get rid of parentheses

4x^2-20x-32x+160-1=0

We add all the numbers together, and all the variables

4x^2-52x+159=0

a = 4; b = -52; c = +159;
Δ = b2-4ac
Δ = -522-4·4·159
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{10}}{2*4}=\frac{52-4\sqrt{10}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{10}}{2*4}=\frac{52+4\sqrt{10}}{8}
$